> timeout = ROUND_UP((unsigned long) timeout, 1000/HZ);
>
> Otherwise, the code is Just Right.
Um, this works perfectly when HZ == 100, but consider what happens when
HZ == 1024. 1000/HZ == 0, and then computing (x+0-1)/0 doesn't work so well.
If you want accuracy with no danger of overflow, try the following trick:
ticks = (msec/1000)*HZ + (msec%1000)*HZ/1000.
To make this more efficient, use that only on large values of msec,
and the simpler (msec*HZ)/1000. In thhe HZ > 1000 case, you also lose
the guarantee that every legal msec value corresponds to a
(C experts will note that none of the parens are necessary, but I though
it was clearer to include them.)
So, for perfection, you want:
unsigned long msec_to_ticks(unsigned long msec)
{
if (msec <= ULONG_MAX/HZ)
return msec*HZ/1000;
#if HZ > 1000
/* Wups, can overflow - saturate return value */
if (msec > (ULONG_MAX/HZ)*1000 + (ULONG_MAX % HZ)*1000/HZ)
return ULONG_MAX
#endif
return (msec/1000)*HZ + (msec%1000)*HZ/1000;
}
Um... this is the rounding-down case, and also saturates at ULONG_MAX
instead of MAX_SCHEDULE_TIMEOUT (LONG_MAX). Let me adjust the boundary
cases a bit...
#if MAX_SCHEDULE_TIMEOUT != LONG_MAX
#error Adjust this code - it assumes identical input and output ranges
#endif
unsigned long msec_to_ticks(unsigned long msec)
{
if (msec < ULONG_MAX/HZ - 999)
return (msec+999)*HZ/1000;
msec--; /* Following code rounds up and adds one */
#if HZ > 1000
/* Wups, can overflow - saturate return value */
if (msec >= (ULONG_MAX/HZ)*1000 + (ULONG_MAX % HZ)*1000/HZ)
return MAX_SCHEDULE_TIMEOUT;
#endif
return (msec/1000)*HZ + (msec%1000)*HZ/1000 + 1;
}
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